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4 years ago

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Problem PageQuestion A cylinder measuring 2.3cm wide and 2.7cm high is filled with gas. The piston is pushed down with a steady

force measured to be 36.N. Calculate the pressure of the gas inside the cylinder. Write your answer in units of kilopascals. Round your answer to 2 significant digits.

1 answer:

Alexandra [31]4 years ago

8 0

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how much energy is needed to raise the temperature of 2kg of copper from 0°c to 10°c. the specific heat capacity of copper is 38

kogti [31]

Explanation:

H= mass× specific heat capacity×change in temperature

=2×380×(10-0)

=2×380×10

=7600Joules

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3 years ago

3. Predict which of the following ions would be likely to form: _____ a. Na2+ _____ b. Cl+ _____ c. Ca2+ _____ d. Br– _____ e. N

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The greater the force exerted by the engine of a car over a set distance, the greater the change in ?

aleksandrvk [35]

The greater the change in velocity, or speed.

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4 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

juin [17]

Answer:

Explanation:

Given

For first case

launch angle $\theta =45^{\circ}C$

at highest point $h=150 m/s$

$150=u\cos 45$

$u=\frac{150}{\cos 45}=212.132 m/s$

For second case

$\theta _2=37^{\circ}C$

at highest Point velocity is $u\cos \theta _2$

$=212.132\times \cos 37$

$=169.41 m/s$

as there is no acceleration in x direction therefore horizontal velocity is same

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3 years ago

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A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

Amanda [17]

A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s$

D) 5.6 rad

The angular displacement of the wheel is given by the equation

$\omega_f^2 - \omega_i^2 = 2 \alpha \theta$

where we have

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel

$\omega_f = 4.00 rad/s$ is the final angular velocity

$\alpha=+1.67 rad/s^2$ is the angular acceleration

Solving for $\theta$ ,

$\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad$

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3 years ago